Question

The energy difference between the ground state of an atom and its excited state is $3\times {10}^{-19}J$. What is the wavelength of the photon required for this transition?

• A. $494.5\times {10}^{-6}m$
• B. $3\times {10}^{-8}m$
• C. $1.8\times {10}^{-7}m$
• D. $6.6\times {10}^{-7}m$

Answer 1

Answer: (D)

$6.6\times {10}^{-7}m$

$\Delta E=E_2-E_1=hc/\lambda$

where, $h=6.626\times {10}^{-34}Js,c=3\times {10}^{8}m{s}^{-1}$

Given $\Delta E=3\times {10}^{-19}J/atom$

$\lambda =6.626\times {10}^{-34}\times 3\times {10}^8/(3\times {10}^{-19})$

$\lambda =6.626\times {10}^{-7}m$

option D